【动态规划】字符串距离

#include<bits/stdc++.h>
using namespace std;
const int N=3000;
char c1[N],c2[N];
int l1,l2,d[N][N],k;
int main() {
    scanf("%s", c1);
    scanf("%s", c2);
    l1 = strlen(c1);
    l2 = strlen(c2);
    scanf("%d", &k);
    for (int i = 1; i <= l1; i++) {
        d[i][0] = d[i - 1][0] + k;
    }
    for (int i = 1; i <= l2; i++) {
        d[0][i] = d[0][i - 1] + k;
    }
    for (int i = 1; i <= l1; i++) {
        for (int j = 1; j <= l2; j++) {
            d[i][j] = min(d[i - 1][j] + k, d[i][j - 1] + k);
            d[i][j] = min(d[i][j], d[i - 1][j - 1] + abs(c1[i - 1] - c2[j - 1]));
        }
    }
    printf("%d
", d[l1][l2]);
}